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We are asked to find the slope of the secant line between \( x_1 = 0 \) and \( x_2 = 4 \) for the given function:

\[
f(x) = 1.3x^3 - 33x^2 + 262x + 559
\]

The formula for the slope of the secant line between two points \( x_1 \) and \( x_2 \) is:

\[
\text{slope of secant} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}
\]

### Step 1: Calculate \( f(0) \)
Substitute \( x = 0 \) into the function:

\[
f(0) = 1.3(0)^3 - 33(0)^2 + 262(0) + 559 = 559
\]

### Step 2: Calculate \( f(4) \)
Substitute \( x = 4 \) into the function:

\[
f(4) = 1.3(4)^3 - 33(4)^2 + 262(4) + 559
\]

Now, I will calculate the exact value of \( f(4) \).We have the following values:

- \( f(0) = 559 \)
- \( f(4) = 1162.2 \)

Now, the slope of the secant line between \( x_1 = 0 \) and \( x_2 = 4 \) is:

\[
\text{slope} = \frac{1162.2 - 559}{4 - 0} = \frac{603.2}{4} = 150.8
\]

Rounding to the nearest whole number, the slope of the secant line is approximately **151**.

Would you like any additional details or explanations? Here are some related questions:

1. How would the slope change if the interval was different, for example, from \( x_1 = 2 \) to \( x_2 = 5 \)?
2. What is the meaning of a secant line in the context of this graph?
3. How can you find the instantaneous rate of change at a particular point, such as at \( x = 4 \)?
4. How would the graph change if the cubic term in \( f(x) \) were modified?
5. Can we approximate the function's maximum or minimum values using its derivatives?

**Tip:** The secant line approximates the average rate of change over a specific interval, while the tangent line represents the instantaneous rate of change at a point.

Find the slope of the secant line from x₁ = 0 to x₂ = 4 for the function f(x) = 1.3x³ - 33x² + 262x + 559.

Learn how to find the slope of the secant line for the cubic function f(x) = 1.3x³ - 33x² + 262x + 559 between x₁ = 0 and x₂ = 4. This problem involves key calculus and algebra concepts, including the secant line slope formula. Suitable for high school students in Grades 11-12.

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